Let \(\varepsilon _{pq}=a+b\sqrt{pq}\)  with \(a, b\in \mathbb {Z}\)  (resp. \(\varepsilon _{2pq}=x+y\sqrt{2pq}\)  with \(x, y\in \mathbb {Z}\)  ) be the fundamental unit of \(\mathbb {Q}(\sqrt{pq})\)  (resp. \(\mathbb {Q}(\sqrt{2pq})\) ). It is known that \(N(\varepsilon _{pq})=N(\varepsilon _{2pq})=1\) . We have \(a^2-1=b^2pq\)  and \(x^2-1=y^22pq\) . So \(a\pm 1\)  and \(x\pm 1\)  are not squares in \(\mathbb {N}\) . In fact, if \(x\pm 1\)  is a square in \(\mathbb {N}\) , then
\( \left\lbrace  \begin{array}{ccc}x\pm 1&=&y_1^2\\x\mp 1&=&2pqy_2^2,\end{array}\right. \) 
for some integers \(y_1\)  and \(y_2\)  such that \(y=y_1y_2\) . So \(1=\left(\frac{y_1^2}{p}\right)=\left(\frac{x\pm 1}{p}\right)=\left(\frac{x\mp 1\pm 2}{p}\right)=\left(\frac{\pm 2}{p}\right)=\left(\frac{ 2}{p}\right)=-1, \)  which is absurd. Similarly \(a\pm 1\)  is not a square in \(\mathbb {N}\) . It follows by [1]} that \(\lbrace \varepsilon _2, \varepsilon _{pq}, \sqrt{\varepsilon _{pq}\varepsilon _{2pq}}\rbrace \)  is a fundamental system of units of \(L_{3, d}^+=\mathbb {Q}(\sqrt{2}, \sqrt{d})\) . Note that by [2]}, we have \(h_2(pq)=h_2(2pq)=2\) .
So by Kuruda's class number formula (see [3]}), we obtain
\(h_2(L_{3, d}^+)=\frac{1}{4}\cdot 2\cdot h_2(pq)h_2(2pq)h_2(2)=2.\) 
Thus \(h_2(d)=h_2(L_{3, d}^+)=2\) . So the result by Theorem REF .
